# Criteria for $3 \times 3$ matrix to positive definite

Here it is said that a $$2\times 2$$ matrix $$A$$ is positive definite if and only if $$tr(A) >0$$ and $$det(A)>0$$. This will not work if $$A$$ is $$3\times 3$$. But is there any way to enforce the positive definiteness of the matrix $$A$$ via the trace and determinant of $$A$$, if $$A$$ is of size $$3\times 3$$?

• Sylvester's Criterion is the most natural extension of this. Note $\det(A) > 0$ implies both eigenvalues are of the same sign, while $\operatorname{tr}(A) > 0$ implies they must both be positive, or equivalently, the top left $1 \times 1$ matrix has positive determinant. This is Sylvester's Criterion for $2 \times 2$ matrices. Note: you also need to assume that the matrix is Hermitian! May 16, 2022 at 15:42

It is possible to consider the Routh-Hurwitz criterion: A matrix $$A$$ is Hurwitz stable if its eigenvalues have negative real part. Moreover, a $$3\times3$$ matrix $$A$$ with characteristic polynomial $$p_A(x)=x^3+p_2x^2+p_1x+p_0$$

is Hurwitz stable if and only if $$p_2>0$$, $$p_0>0$$ and $$p_0-p_1p_2<0$$.

So, a symmetric matrix $$A$$ is positive definite if and only if $$-A$$ is Hurwitz stable.

The characteristic polynomial of $$A$$ is given by

$$p_A(x)=x^3-\mathrm{trace}(A)x^2+\mathrm{trace}(\mathrm{Adj}(A))x-\det(A)$$

where $$\mathrm{Adj}(A)$$ is the adjugate matrix of $$A$$. Consequently, the characteristic polynomial of $$-A$$ is given by

$$p_{-A}(x)=x^3+\mathrm{trace}(A)x^2-\mathrm{trace}(\mathrm{Adj}(A))x+\det(A).$$

The Routh-Hurwitz criterion for the stability of $$-M$$ is given by

• $$\mathrm{trace}(A)>0$$,
• $$\det(A)>0$$,
• $$\det(A)+\mathrm{trace}(\mathrm{Adj}(A))\mathrm{trace}(A)>0$$.
• N.b. that $\operatorname{trace}(\operatorname{Adj}(-A)) = \operatorname{trace}(\operatorname{Adj}(A))$. May 16, 2022 at 16:24
• @TravisWillse Thanks, I will update the answer.
– KBS
May 16, 2022 at 16:33
• Incidentally, I think we can simplify the criterion using the fact that, a priori, the eigenvalues of a symmetric matrix are real; see my answer. May 16, 2022 at 16:56

No, it is not possible. Take any two real numbers $$a$$ and $$b$$ such that $$a^2+b^2=1$$ and the matrix $$A= \begin{pmatrix} 1 & a & b \\ a & a^2+1 & a b \\ b & a b & -a^2+2 \end{pmatrix}.$$ Then $$\operatorname{tr}(A)=4$$, $$\det(A)=1$$, and $$A$$ is definite positive.

But if you take two real numbers $$a$$ and $$b$$ such that $$a^2+b^2=5$$ and you define $$B= \begin{pmatrix} 1&a&b\\ a&a^2-1&ab\\ b&ab&-a^2+4 \end{pmatrix},$$ then $$\operatorname{tr}(B)=4$$, $$\det(B)=1$$, but $$B$$ is not definite positive.

• How did you find these examples? May 16, 2022 at 17:05
• I took a $3\times3$ symmetric matrix$$M=\begin{pmatrix}1&a&b\\a&c&d\\b&c&e\end{pmatrix}$$and then I found a solution of the system$$\left\{\begin{array}{l}\operatorname{tr}(M)=4\\c-a^2=1\\\det(M)=1\end{array}\right.$$That's how I got $A$. In order to get $B$, I did almost the same thing, the only difference being that I used the equation $c-a^2=-1$ instead. May 16, 2022 at 17:11

No, positive definiteness of a (symmetric) $$3 \times 3$$ matrix $$A$$ cannot be determined using $$\det A$$ and $$\operatorname{tr} A$$ alone.

(Counter)example The diagonal matrix $$\pmatrix{-\frac{1}{2}\\&-\frac{1}{2}\\&&4}$$ has the same trace ($$3$$) and determinant ($$1$$) as the (positive definite) $$3 \times 3$$ identity matrix but is not positive definite.

In any case we can generalize the criterion for a $$2 \times 2$$ matrix to $$3 \times 3$$---in fact $$n \times n$$---matrices.

Denote the eigenvalues $$A$$ by $$\lambda, \mu, \nu$$; since $$A$$ is symmetric, all $$3$$ are real. The characteristic polynomial of $$A$$ is $$p_A(t) = t^3 - (\lambda + \mu + \nu) t^2 + (\mu \nu + \nu \lambda + \lambda \mu) t - \lambda \mu \nu,$$ or, more compactly, $$p_A(t) = t^3 - \operatorname{tr} A \cdot t^2 + \sigma_2(A) t - \det A ,$$ where $$\sigma_2(A) := \mu \nu + \nu \lambda + \lambda \mu$$, i.e., the second elementary symmetric polynomial in the eigenvalues of $$A$$. So, if $$\operatorname{tr} A, \sigma_2(A), \det A > 0$$, Descartes' Rule of Signs implies that the (again, real) roots $$\lambda, \mu, \nu$$ of $$p_A$$ are positive, equivalently that $$A$$ is positive definite. Conversely, if $$A$$ is positive definite, $$\sigma_2(A) = \mu \nu + \nu \lambda + \lambda \mu > 0$$. Thus, $$\boxed{\textrm{A is positive definite} \qquad \textrm{iff} \qquad \left\{\begin{array}{r} \operatorname{tr} A > 0 \\ \sigma_2(A) > 0 \\ \operatorname{det} A > 0 \end{array}\right.} .$$ The same reasoning immediately yields a generalization to (symmetric) matrices of arbitrary size: $$\boxed{\textrm{A is positive definite} \qquad \textrm{iff} \qquad (\sigma_k(A) = 0 \qquad \textrm{for all } k \in \{1, \ldots, n\})} ,$$ where $$\sigma_k(A)$$ is the $$i$$th elementary symmetric polynomial in the eigenvalues $$\lambda_1, \ldots, \lambda_n$$ of $$A$$, $$\sigma_k(A) := \sum_{1 \leq i_1 < \cdots i_k \leq n} \lambda_{i_1} \cdots \lambda_{i_k} .$$ Notice that $$\sigma_1 = \operatorname{tr}$$ and $$\sigma_n = \operatorname{det}$$, so this statement specializes to the result for $$n = 2$$ given in the question statement.

Remark We have the identity $$\sigma_2 = \operatorname{tr} \circ \operatorname{Adj}$$, where $$\operatorname{Adj} A$$ denotes the adjugate (a.k.a., confusingly, the classical adjoint) of $$A$$. We can also express $$\sigma(A)$$ in terms of traces of $$A, A^2$$: $$\sigma_2(A) = \frac{1}{2}\left[(\operatorname{tr} A)^2 - \operatorname{tr} (A^2)\right].$$

• I upvoted but I still find it too complex. May 16, 2022 at 17:00
• Actually nvm this is great, I don't know if using Rule of signs is necessary but w/e . It might also be good mentioning how the dim $2$ analogue is basically just doing this but easier. May 16, 2022 at 17:09
• @Asinomás Thanks for the comment. I added comments to the end of the answer explaining more explicitly how the generalization to the $n \times n$ case works and that the general method specializes to the result OP mentioned for $n = 2$. May 16, 2022 at 17:23
• And you're right, we don't need the full strength of Descartes' Rule of Signs; it just shortens the presentation some. All we need is that the alternation of the signs of $p_A(t)$ means that the coefficients of $p_A(-t)$ all have the same sign, hence $p_A(-t)$ has no positive roots, or equivalently that $p_A$ has no negative roots. May 16, 2022 at 17:28